# How do you divide (4+2i )/( 1-i)?

Feb 8, 2016

$1 + 3 i$

#### Explanation:

You must eliminate the complex number in the denominator by multiplying by its conjugate:

$\frac{4 + 2 i}{1 - i} = \frac{\left(4 + 2 i\right) \left(1 + i\right)}{\left(1 - i\right) \left(1 + i\right)}$

$\frac{4 + 4 i + 2 i + 2 {i}^{2}}{1 - {i}^{2}}$

$\frac{4 + 6 i - 2}{1 + 1}$

$\frac{2 + 6 i}{2}$

$1 + 3 i$

Feb 8, 2016

1 + 3i

#### Explanation:

Require the denominator to be real . To achieve this multiply the numerator and the denominator by the complex conjugate of the denominator.

If (a + bi ) is a complex number then (a - bi) is the conjugate

here the conjugate of (1 - i ) is (1 + i )

now $\frac{\left(4 + 2 i\right) \left(1 + i\right)}{\left(1 - i\right) \left(1 + i\right)}$

distribute the brackets to obtain :

$\frac{4 + 6 i + 2 {i}^{2}}{1 - {i}^{2}}$

note that ${i}^{2} = \left({\sqrt{- 1}}^{2}\right) = - 1$

hence $\frac{4 + 6 i - 2}{1 + 1} = \frac{2 + 6 i}{2} = \frac{2}{2} + \frac{6 i}{2} = 1 + 3 i$